172283
The equations of displacement of two waves are $y_{1}=10 \sin (2 \pi t+\pi / 3)$ and $y_{2}=5[\sin 3 \pi t+$ $\sqrt{3} \cos 3 \pi t]$. What is the ratio of their amplitude?
1 $1: 2$
2 $2: 1$
3 $1: 1$
4 None of the above
Explanation:
C Given that, $\mathrm{y}_{1}=10 \sin \left(2 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude to this wave is $\mathrm{A}_{1}=10$ And $\mathrm{y}_{2}=5[\sin 3 \pi \mathrm{t}+\sqrt{3} \cos 3 \pi \mathrm{t}] \times \frac{2}{2}$ $=10\left[\frac{1}{2} \sin 3 \pi \mathrm{t}+\frac{\sqrt{3}}{2} \cos 3 \pi \mathrm{t}\right]$ $=10\left[\cos \frac{\pi}{3} \sin 3 \pi \mathrm{t}+\sin \frac{\pi}{3} \cos 3 \pi \mathrm{t}\right]$ $=10 \sin \left(3 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude of this wave is $\mathrm{A}_{2}=10$ Then, $\frac{A_{1}}{A_{2}}=\frac{10}{10}=1: 1$
UP CPMT-2012
WAVES
172284
Equation of a progressive wave is given by $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ the distance is expressed in $\mathrm{cm}$ and time in second. What will be the minimum distance between two particles having the phase difference of $\frac{\pi}{2}$ ?
1 $4 \mathrm{~cm}$
2 $8 \mathrm{~cm}$
3 $25 \mathrm{~cm}$
4 $12.5 \mathrm{~cm}$
Explanation:
C Equation of a progressive wave is given by - $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ $y=0.2 \cos \left(0.04 \pi t+0.02 \pi x-\pi^{2} / 6\right)$ Comparing by general equation of wave progression- $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx}-\phi)$ We get, $\omega=0.04 \pi$ $\mathrm{k}=0.02 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}=0.02 \pi$ $\lambda=\frac{2 \pi}{0.02 \pi}=100 \mathrm{~cm}$ $\Delta \phi=\pi / 2$ Hence path difference between them $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}=\frac{100}{2 \pi} \times \frac{\pi}{2}=\frac{100}{4}$ $\Delta \mathrm{x}=25 \mathrm{~cm}$
UP CPMT-2011
WAVES
172287
The equation of a stationary wave is given by $y$ $=0.4 \sin 160 \pi t \cos \frac{\pi}{16} x$, where $t$ is in second, $x$ and $y$ in $\mathrm{cm}$. Separation between successive nodes is
1 $32 \mathrm{~cm}$
2 $16 \mathrm{~cm}$
3 $8 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Standard equation of stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{T}} \cos \frac{2 \pi \mathrm{x}}{\lambda}$ Given equation $\mathrm{y}=0.4 \sin 160 \pi \mathrm{t} \cos \frac{\pi \mathrm{x}}{16}$ On comparing equation (i) with (ii) $\frac{2 \pi}{\lambda}=\frac{\pi}{16}$ $\lambda=32 \mathrm{~cm}$ Distance between two successive nodes $=\frac{\lambda}{2}=\frac{32}{2}=16 \mathrm{~cm}$
UP CPMT-2001
WAVES
172290
The equation of a transverse wave is given by $y$ $=10 \sin \pi(0.01 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in seconds. Its frequency is
172283
The equations of displacement of two waves are $y_{1}=10 \sin (2 \pi t+\pi / 3)$ and $y_{2}=5[\sin 3 \pi t+$ $\sqrt{3} \cos 3 \pi t]$. What is the ratio of their amplitude?
1 $1: 2$
2 $2: 1$
3 $1: 1$
4 None of the above
Explanation:
C Given that, $\mathrm{y}_{1}=10 \sin \left(2 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude to this wave is $\mathrm{A}_{1}=10$ And $\mathrm{y}_{2}=5[\sin 3 \pi \mathrm{t}+\sqrt{3} \cos 3 \pi \mathrm{t}] \times \frac{2}{2}$ $=10\left[\frac{1}{2} \sin 3 \pi \mathrm{t}+\frac{\sqrt{3}}{2} \cos 3 \pi \mathrm{t}\right]$ $=10\left[\cos \frac{\pi}{3} \sin 3 \pi \mathrm{t}+\sin \frac{\pi}{3} \cos 3 \pi \mathrm{t}\right]$ $=10 \sin \left(3 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude of this wave is $\mathrm{A}_{2}=10$ Then, $\frac{A_{1}}{A_{2}}=\frac{10}{10}=1: 1$
UP CPMT-2012
WAVES
172284
Equation of a progressive wave is given by $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ the distance is expressed in $\mathrm{cm}$ and time in second. What will be the minimum distance between two particles having the phase difference of $\frac{\pi}{2}$ ?
1 $4 \mathrm{~cm}$
2 $8 \mathrm{~cm}$
3 $25 \mathrm{~cm}$
4 $12.5 \mathrm{~cm}$
Explanation:
C Equation of a progressive wave is given by - $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ $y=0.2 \cos \left(0.04 \pi t+0.02 \pi x-\pi^{2} / 6\right)$ Comparing by general equation of wave progression- $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx}-\phi)$ We get, $\omega=0.04 \pi$ $\mathrm{k}=0.02 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}=0.02 \pi$ $\lambda=\frac{2 \pi}{0.02 \pi}=100 \mathrm{~cm}$ $\Delta \phi=\pi / 2$ Hence path difference between them $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}=\frac{100}{2 \pi} \times \frac{\pi}{2}=\frac{100}{4}$ $\Delta \mathrm{x}=25 \mathrm{~cm}$
UP CPMT-2011
WAVES
172287
The equation of a stationary wave is given by $y$ $=0.4 \sin 160 \pi t \cos \frac{\pi}{16} x$, where $t$ is in second, $x$ and $y$ in $\mathrm{cm}$. Separation between successive nodes is
1 $32 \mathrm{~cm}$
2 $16 \mathrm{~cm}$
3 $8 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Standard equation of stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{T}} \cos \frac{2 \pi \mathrm{x}}{\lambda}$ Given equation $\mathrm{y}=0.4 \sin 160 \pi \mathrm{t} \cos \frac{\pi \mathrm{x}}{16}$ On comparing equation (i) with (ii) $\frac{2 \pi}{\lambda}=\frac{\pi}{16}$ $\lambda=32 \mathrm{~cm}$ Distance between two successive nodes $=\frac{\lambda}{2}=\frac{32}{2}=16 \mathrm{~cm}$
UP CPMT-2001
WAVES
172290
The equation of a transverse wave is given by $y$ $=10 \sin \pi(0.01 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in seconds. Its frequency is
172283
The equations of displacement of two waves are $y_{1}=10 \sin (2 \pi t+\pi / 3)$ and $y_{2}=5[\sin 3 \pi t+$ $\sqrt{3} \cos 3 \pi t]$. What is the ratio of their amplitude?
1 $1: 2$
2 $2: 1$
3 $1: 1$
4 None of the above
Explanation:
C Given that, $\mathrm{y}_{1}=10 \sin \left(2 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude to this wave is $\mathrm{A}_{1}=10$ And $\mathrm{y}_{2}=5[\sin 3 \pi \mathrm{t}+\sqrt{3} \cos 3 \pi \mathrm{t}] \times \frac{2}{2}$ $=10\left[\frac{1}{2} \sin 3 \pi \mathrm{t}+\frac{\sqrt{3}}{2} \cos 3 \pi \mathrm{t}\right]$ $=10\left[\cos \frac{\pi}{3} \sin 3 \pi \mathrm{t}+\sin \frac{\pi}{3} \cos 3 \pi \mathrm{t}\right]$ $=10 \sin \left(3 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude of this wave is $\mathrm{A}_{2}=10$ Then, $\frac{A_{1}}{A_{2}}=\frac{10}{10}=1: 1$
UP CPMT-2012
WAVES
172284
Equation of a progressive wave is given by $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ the distance is expressed in $\mathrm{cm}$ and time in second. What will be the minimum distance between two particles having the phase difference of $\frac{\pi}{2}$ ?
1 $4 \mathrm{~cm}$
2 $8 \mathrm{~cm}$
3 $25 \mathrm{~cm}$
4 $12.5 \mathrm{~cm}$
Explanation:
C Equation of a progressive wave is given by - $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ $y=0.2 \cos \left(0.04 \pi t+0.02 \pi x-\pi^{2} / 6\right)$ Comparing by general equation of wave progression- $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx}-\phi)$ We get, $\omega=0.04 \pi$ $\mathrm{k}=0.02 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}=0.02 \pi$ $\lambda=\frac{2 \pi}{0.02 \pi}=100 \mathrm{~cm}$ $\Delta \phi=\pi / 2$ Hence path difference between them $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}=\frac{100}{2 \pi} \times \frac{\pi}{2}=\frac{100}{4}$ $\Delta \mathrm{x}=25 \mathrm{~cm}$
UP CPMT-2011
WAVES
172287
The equation of a stationary wave is given by $y$ $=0.4 \sin 160 \pi t \cos \frac{\pi}{16} x$, where $t$ is in second, $x$ and $y$ in $\mathrm{cm}$. Separation between successive nodes is
1 $32 \mathrm{~cm}$
2 $16 \mathrm{~cm}$
3 $8 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Standard equation of stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{T}} \cos \frac{2 \pi \mathrm{x}}{\lambda}$ Given equation $\mathrm{y}=0.4 \sin 160 \pi \mathrm{t} \cos \frac{\pi \mathrm{x}}{16}$ On comparing equation (i) with (ii) $\frac{2 \pi}{\lambda}=\frac{\pi}{16}$ $\lambda=32 \mathrm{~cm}$ Distance between two successive nodes $=\frac{\lambda}{2}=\frac{32}{2}=16 \mathrm{~cm}$
UP CPMT-2001
WAVES
172290
The equation of a transverse wave is given by $y$ $=10 \sin \pi(0.01 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in seconds. Its frequency is
172283
The equations of displacement of two waves are $y_{1}=10 \sin (2 \pi t+\pi / 3)$ and $y_{2}=5[\sin 3 \pi t+$ $\sqrt{3} \cos 3 \pi t]$. What is the ratio of their amplitude?
1 $1: 2$
2 $2: 1$
3 $1: 1$
4 None of the above
Explanation:
C Given that, $\mathrm{y}_{1}=10 \sin \left(2 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude to this wave is $\mathrm{A}_{1}=10$ And $\mathrm{y}_{2}=5[\sin 3 \pi \mathrm{t}+\sqrt{3} \cos 3 \pi \mathrm{t}] \times \frac{2}{2}$ $=10\left[\frac{1}{2} \sin 3 \pi \mathrm{t}+\frac{\sqrt{3}}{2} \cos 3 \pi \mathrm{t}\right]$ $=10\left[\cos \frac{\pi}{3} \sin 3 \pi \mathrm{t}+\sin \frac{\pi}{3} \cos 3 \pi \mathrm{t}\right]$ $=10 \sin \left(3 \pi \mathrm{t}+\frac{\pi}{3}\right)$ $\therefore$ Amplitude of this wave is $\mathrm{A}_{2}=10$ Then, $\frac{A_{1}}{A_{2}}=\frac{10}{10}=1: 1$
UP CPMT-2012
WAVES
172284
Equation of a progressive wave is given by $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ the distance is expressed in $\mathrm{cm}$ and time in second. What will be the minimum distance between two particles having the phase difference of $\frac{\pi}{2}$ ?
1 $4 \mathrm{~cm}$
2 $8 \mathrm{~cm}$
3 $25 \mathrm{~cm}$
4 $12.5 \mathrm{~cm}$
Explanation:
C Equation of a progressive wave is given by - $y=0.2 \cos \pi\left(0.04 t+0.02 x-\frac{\pi}{6}\right)$ $y=0.2 \cos \left(0.04 \pi t+0.02 \pi x-\pi^{2} / 6\right)$ Comparing by general equation of wave progression- $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx}-\phi)$ We get, $\omega=0.04 \pi$ $\mathrm{k}=0.02 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}=0.02 \pi$ $\lambda=\frac{2 \pi}{0.02 \pi}=100 \mathrm{~cm}$ $\Delta \phi=\pi / 2$ Hence path difference between them $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \Delta \phi$ $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}=\frac{100}{2 \pi} \times \frac{\pi}{2}=\frac{100}{4}$ $\Delta \mathrm{x}=25 \mathrm{~cm}$
UP CPMT-2011
WAVES
172287
The equation of a stationary wave is given by $y$ $=0.4 \sin 160 \pi t \cos \frac{\pi}{16} x$, where $t$ is in second, $x$ and $y$ in $\mathrm{cm}$. Separation between successive nodes is
1 $32 \mathrm{~cm}$
2 $16 \mathrm{~cm}$
3 $8 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Standard equation of stationary wave equation $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{T}} \cos \frac{2 \pi \mathrm{x}}{\lambda}$ Given equation $\mathrm{y}=0.4 \sin 160 \pi \mathrm{t} \cos \frac{\pi \mathrm{x}}{16}$ On comparing equation (i) with (ii) $\frac{2 \pi}{\lambda}=\frac{\pi}{16}$ $\lambda=32 \mathrm{~cm}$ Distance between two successive nodes $=\frac{\lambda}{2}=\frac{32}{2}=16 \mathrm{~cm}$
UP CPMT-2001
WAVES
172290
The equation of a transverse wave is given by $y$ $=10 \sin \pi(0.01 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in seconds. Its frequency is
